Simultaneous linear & quadratic equations

1. Revise simultaneous linear equations


ID is: 3270 Seed is: 4961

Simultaneous equations: the elimination method

Here are two equations, which we can solve simultaneously using elimination:

3x3y=125x+3y=14

If we add these equations together, which of the terms will be eliminated (cancelled), and why does the elimination happen? Choose your answers from the choices below.

Answer:

We will eliminate because .

2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?


STEP: Identify the terms which are easy to cancel
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two x-terms and two y-terms:

3x3y=125x+3y=14

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the x-terms or the y-terms.

For the equations in this question, notice that the coefficients of y are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

3x3y=125x+3y=143x5x3y+3y=1214disappear!y-terms 2x+0y=22x=2

The y-terms cancel each other out - that is what 'elimination' means! We get the 0y because the y-coefficients are equal and opposite: 3y+(3y) is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the y-terms because the y-coefficients will cancel.


Submit your answer as: and

ID is: 3270 Seed is: 7122

Simultaneous equations: the elimination method

Here are two equations, which we can solve simultaneously using elimination:

3x4y=13x2y=5

If we add these equations together, which of the terms will be eliminated (cancelled), and why does the elimination happen? Choose your answers from the choices below.

Answer:

We will eliminate because .

2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?


STEP: Identify the terms which are easy to cancel
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two x-terms and two y-terms:

3x4y=13x2y=5

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the x-terms or the y-terms.

For the equations in this question, notice that the coefficients of x are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

3x4y=13x2y=53x+3x4y2y=15disappear!x-terms 0x6y=66y=6

The x-terms cancel each other out - that is what 'elimination' means! We get the 0x because the x-coefficients are equal and opposite: 3x+(3x) is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the x-terms because the x-coefficients will cancel.


Submit your answer as: and

ID is: 3270 Seed is: 1600

Simultaneous equations: the elimination method

Here are two equations, which we can solve simultaneously using elimination:

4=3x+y1=3x+4y

If we add these equations together, which of the terms will be eliminated (cancelled), and why does the elimination happen? Choose your answers from the choices below.

Answer:

We will eliminate because .

2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?


STEP: Identify the terms which are easy to cancel
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two x-terms and two y-terms:

4=3x+y1=3x+4y

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the x-terms or the y-terms.

For the equations in this question, notice that the coefficients of x are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

4=3x+y1=3x+4y41=3x3x+y+4ydisappear!x-terms 5=0x+5y5=5y

The x-terms cancel each other out - that is what 'elimination' means! We get the 0x because the x-coefficients are equal and opposite: 3x+(3x) is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the x-terms because the x-coefficients will cancel.


Submit your answer as: and

ID is: 379 Seed is: 5385

Solving simultaneous equations by elimination

Solve for m and n:

12m+2n=06m+3n=12
Answer: m= and n=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the m-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Multiply one of the equations to set up the cancellation
[−2 points ⇒ 4 / 6 points left]

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the m-term in the first equation is a multiple of the second m-term: the 6 is 2 times as much as 12. If we multiply the entire second equation by this factor of 2 we will be able to cancel the m-terms between the equations:

6m+3n=12(2)6m+(2)3n=12(2)12m+6n=24
NOTE: We could also multiply the equation by −2. That will also make it possible to cancel the m-terms (it will just result in opposite signs for all of the coefficients).

STEP: Eliminate m by subtracting the equations
[−2 points ⇒ 2 / 6 points left]

Now eliminate the m-terms. Since the terms have the same signs, we must subtract one equation from the other to cancel those terms.

12m+2n=0to cancel the m-termsSubtract the equations(12m+6n)=240m4n=24

Now we can solve for n:

n=244=6

STEP: Solve for m
[−2 points ⇒ 0 / 6 points left]

Finally, use the value of n to find the value of m. Remember that we can use either of the equations to do this, we can pick the easier equation. The equations are pretty much the same - for example, neither has any fractions or negatives. So we will use the first one.

12m+2n=012m+2(6)=012m+12=012m=12m=1

The answers to the pair of equations are m=1 and n=6.


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ID is: 379 Seed is: 4931

Solving simultaneous equations by elimination

Solve for f and g:

12f+6g=363f+4g=2
Answer: f= and g=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the f-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Multiply one of the equations to set up the cancellation
[−2 points ⇒ 4 / 6 points left]

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the f-term in the first equation is a multiple of the second f-term: the 3 is 4 times as much as 12. If we multiply the entire second equation by this factor of 4 we will be able to cancel the f-terms between the equations:

3f+4g=2(4)3f+(4)4g=2(4)12f+16g=8
NOTE: We could also multiply the equation by −4. That will also make it possible to cancel the f-terms (it will just result in opposite signs for all of the coefficients).

STEP: Eliminate f by adding the equations
[−2 points ⇒ 2 / 6 points left]

Now eliminate the f-terms. Since the terms have opposite signs, we must add one equation to the other to cancel those terms.

12f+6g=36to cancel the f-termsAdd the equations+(12f+16g)=+80f+22g=44

Now we can solve for g:

g=4422=2

STEP: Solve for f
[−2 points ⇒ 0 / 6 points left]

Finally, use the value of g to find the value of f. Remember that we can use either of the equations to do this, we can pick the easier equation. The second equation looks like a better choice because it has fewer negative signs than the first equation (in fact it does not have any!).

3f+4g=23f+4(2)=23f+8=23f=6f=2

The answers to the pair of equations are f=2 and g=2.


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ID is: 379 Seed is: 3159

Solving simultaneous equations by elimination

Solve for j and k:

2j8k=183j4k=5
Answer: j= and k=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the k-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Multiply one of the equations to set up the cancellation
[−2 points ⇒ 4 / 6 points left]

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the k-term in the first equation is a multiple of the second k-term: the 4 is 2 times as much as 8. If we multiply the entire second equation by this factor of 2 we will be able to cancel the k-terms between the equations:

3j4k=5(2)3j(2)4k=5(2)6j8k=10
NOTE: We could also multiply the equation by −2. That will also make it possible to cancel the k-terms (it will just result in opposite signs for all of the coefficients).

STEP: Eliminate k by subtracting the equations
[−2 points ⇒ 2 / 6 points left]

Now eliminate the k-terms. Since the terms have the same signs, we must subtract one equation from the other to cancel those terms.

2j8k=18to cancel the k-termsSubtract the equations(6j8k)=104j+0k=28

Now we can solve for j:

j=284=7

STEP: Solve for k
[−2 points ⇒ 0 / 6 points left]

Finally, use the value of j to find the value of k. Remember that we can use either of the equations to do this, we can pick the easier equation. The second equation looks like a better choice because it has fewer negative signs than the first equation.

3j4k=53(7)4k=5214k=54k=16k=4

The answers to the pair of equations are j=7 and k=4.


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ID is: 3265 Seed is: 9522

Simultaneous equations

Solve the following equations simultaneously:

xy=85x+y=20
Answer:

The numbers which solve both equations are x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Notice that the y-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.


STEP: Eliminate (cancel) the y-terms
[−1 point ⇒ 2 / 3 points left]

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for x and for y, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the y-coefficients are equal and opposite. If we add the equations together, the y-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

xy=85x+y=20x+5xy+y=8204x+0y=124x=12

STEP: Solve for x
[−1 point ⇒ 1 / 3 points left]

Now we have an equation with only one variable, and we can solve it.

4x=12x=3

STEP: Find the value of y
[−1 point ⇒ 0 / 3 points left]

So x=3. But we are not done yet: we also need an answer for y. We can get this using the answer we just got for x. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the second equation, because it has fewer negative signs.

5x+y=205(3)+y=2015+y=20y=5

Now we have the complete answer: the numbers which solve the equations are x=3 and y=5. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are x=3 and y=5.


Submit your answer as: and

ID is: 3265 Seed is: 340

Simultaneous equations

Solve the following equations simultaneously:

14=x4y6=3x+4y
Answer:

The numbers which solve both equations are x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Notice that the y-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.


STEP: Eliminate (cancel) the y-terms
[−1 point ⇒ 2 / 3 points left]

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for x and for y, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the y-coefficients are equal and opposite. If we add the equations together, the y-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

14=x4y6=3x+4y14+6=x+3x4y+4y8=4x+0y8=4x

STEP: Solve for x
[−1 point ⇒ 1 / 3 points left]

Now we have an equation with only one variable, and we can solve it.

8=4x2=x

STEP: Find the value of y
[−1 point ⇒ 0 / 3 points left]

So x=2. But we are not done yet: we also need an answer for y. We can get this using the answer we just got for x. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the second equation, because it has no negative signs.

6=3x+4y6=3(2)+4y6=6+4y12=4y3=y

Now we have the complete answer: the numbers which solve the equations are x=2 and y=3. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are x=2 and y=3.


Submit your answer as: and

ID is: 3265 Seed is: 1785

Simultaneous equations

Solve the following equations simultaneously:

2x+4y=105x4y=19
Answer:

The numbers which solve both equations are x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Notice that the y-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.


STEP: Eliminate (cancel) the y-terms
[−1 point ⇒ 2 / 3 points left]

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for x and for y, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the y-coefficients are equal and opposite. If we add the equations together, the y-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

2x+4y=105x4y=192x+5x+4y4y=10193x+0y=93x=9

STEP: Solve for x
[−1 point ⇒ 1 / 3 points left]

Now we have an equation with only one variable, and we can solve it.

3x=9x=3

STEP: Find the value of y
[−1 point ⇒ 0 / 3 points left]

So x=3. But we are not done yet: we also need an answer for y. We can get this using the answer we just got for x. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the first equation, because it has fewer negative signs.

2x+4y=102(3)+4y=106+4y=104y=4y=1

Now we have the complete answer: the numbers which solve the equations are x=3 and y=1. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are x=3 and y=1.


Submit your answer as: and

ID is: 3288 Seed is: 7154

Solving simultaneous equations

Solve the following equations simultaneously. Use whichever method is easiest.

4x+5y=115y=x16
Answer:

The solution is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.


STEP: Select the method for solving the question
[−1 point ⇒ 4 / 5 points left]

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations invite elimination, because the y-terms have equal coefficients. We can eliminate the y-terms by subtracting the equations.

NOTE: We can solve these equations using substitution - it will work. But it will probably be more complicated than elimination.

STEP: Do the elimination and solve for x
[−3 points ⇒ 1 / 5 points left]

Eliminate the y and solve for x:

4x+5y=11second equationSubtract the(5y)=(x16)4x5y+5y=11+16x4x=x+273x=27x=9

Super - we know that x=9.


STEP: Determine the other variable
[−1 point ⇒ 0 / 5 points left]

Now we can use the x-value to find the value of y. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case the equations are pretty similar, so we will just pick the first one.

4x+5y=114(9)+5y=115y=25y=5

The answer is the pair of numbers x=9 and y=5. On the Cartesian plane, this looks like this:

The values which solve these two equations are x=9 and y=5.


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ID is: 3288 Seed is: 2178

Solving simultaneous equations

Solve the following equations simultaneously. Use whichever method is easiest.

5y=2x152y=3x13
Answer:

The solution is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.


STEP: Select the method for solving the question
[−1 point ⇒ 5 / 6 points left]

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations are not arranged nicely for elimination or substitution. And it is not possible to make a simple change to these equations. We need to multiply both of the equations so that we can get terms which will cancel. Based on the coefficients, one choice is to multiply the first equation by 3 and the second equation by 2. Then we can cancel the x-terms.

NOTE: We can solve these equations using substitution - it will work. But it will probably be more complicated than elimination.

STEP: Do the elimination and solve for y
[−4 points ⇒ 1 / 6 points left]

First change the first and second equation to set up the elimination:

3(5y)=3(2x15)15y=6x45

and

2(2y)=2(3x13)4y=6x26

Now do the elimination and complete the solution to find y.

15y=6x45second equationSubtract the(4y)=(6x26)15y4y=45+266x+6x19y=19y=1

Super - we know that y=1.


STEP: Determine the other variable
[−1 point ⇒ 0 / 6 points left]

Now we can use the y-value to find the value of x. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

2y=3x132(1)=3x133x=15x=5

The answer is the pair of numbers x=5 and y=1. On the Cartesian plane, this looks like this:

The values which solve these two equations are x=5 and y=1.


Submit your answer as: and

ID is: 3288 Seed is: 6498

Solving simultaneous equations

Solve the following equations simultaneously. Use whichever method is easiest.

5y10=2xx2y=5
Answer:

The solution is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.


STEP: Select the method for solving the question
[−1 point ⇒ 5 / 6 points left]

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to subtract the 2y term to the other side of the second equation.

NOTE: We can solve these equations using elimination - it will work. But it will probably be more complicated than substitution.

STEP: Do the substitution and solve for y
[−4 points ⇒ 1 / 6 points left]

First rearrange the second equation to set up the substitution:

x2y=5x=2y+5

Now do the substitution and complete the solution to find y.

5y10=2x5y10=2(2y+5)5y10=4y109y=0y=0

Super - we know that y=0.


STEP: Determine the other variable
[−1 point ⇒ 0 / 6 points left]

Now we can use the y-value to find the value of x. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

x2y=5x2(0)=5x=5

The answer is the pair of numbers x=5 and y=0. On the Cartesian plane, this looks like this:

The values which solve these two equations are x=5 and y=0.


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ID is: 389 Seed is: 9038

Solving simultaneous equations by substitution

Solve for x and y using substitution:

9x8y=9and5x3y=5

Your answers should be exact (do not round off).

Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.


STEP: Prepare one of the equations for substitution
[−1 point ⇒ 3 / 4 points left]

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of 1 or 1, because that would make our work easier. But that is not the case. So we will just make x the subject of the first equation:

9x8y=99x=8y+9x=8y91

Unfortunately, this equation has 1 fraction in it. That is ugly, but it will work.


STEP: Substitute and solve for y
[−2 points ⇒ 1 / 4 points left]

Substitute the expression for x into the second equation. Then simplify and solve for the value of y:

5x3y=5in for xsubstitute5(8y91)3y=513y9+5=513y9=10y=9013

STEP: Find the value of x
[−1 point ⇒ 0 / 4 points left]

Substitute this value of y back into one of the equations and solve for x. In this case, the first equation has only 2 negative signs while the other equation has 3 negatives. So we will use the first equation.

9x8y=99x8(9013)=99x=60313x=6713

The final answers are x=6713 and y=9013.


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ID is: 389 Seed is: 4973

Solving simultaneous equations by substitution

Solve for x and y using substitution:

2x+3y=10and6x9y=9

Your answers should be exact (do not round off).

Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.


STEP: Prepare one of the equations for substitution
[−1 point ⇒ 3 / 4 points left]

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of 1 or 1, because that would make our work easier. But that is not the case. So we will just make x the subject of the first equation:

2x+3y=102x=3y10x=3y25

Unfortunately, this equation has 1 fraction in it. That is ugly, but it will work.


STEP: Substitute and solve for y
[−2 points ⇒ 1 / 4 points left]

Substitute the expression for x into the second equation. Then simplify and solve for the value of y:

6x9y=9in for xsubstitute6(3y25)9y=918y30=918y=39y=136

STEP: Find the value of x
[−1 point ⇒ 0 / 4 points left]

Substitute this value of y back into one of the equations and solve for x. In this case, the equations are similar: they both have 1 negative sign. We will pick the first equation.

2x+3y=102x+3(136)=102x=72x=74

The final answers are x=74 and y=136.


Submit your answer as: and

ID is: 389 Seed is: 7158

Solving simultaneous equations by substitution

Solve for x and y using substitution:

3x+5y=5and9x3y=0

Your answers should be exact (do not round off).

Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.


STEP: Prepare one of the equations for substitution
[−1 point ⇒ 3 / 4 points left]

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of 1 or 1, because that would make our work easier. But that is not the case. So we will just make x the subject of the first equation:

3x+5y=53x=5y+5x=5y3+53

Unfortunately, this equation has 2 fractions in it. That is ugly, but it will work.


STEP: Substitute and solve for y
[−2 points ⇒ 1 / 4 points left]

Substitute the expression for x into the second equation. Then simplify and solve for the value of y:

9x3y=0in for xsubstitute9(5y3+53)3y=012y15=012y=15y=54

STEP: Find the value of x
[−1 point ⇒ 0 / 4 points left]

Substitute this value of y back into one of the equations and solve for x. In this case, we should use the first equation because it has no negative signs.

3x+5y=53x+5(54)=53x=54x=512

The final answers are x=512 and y=54.


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ID is: 3289 Seed is: 1342

Equations with fractions

Solve the following equations simultaneously to find the values of j and k:

2j=2k+10jkj,k04k=3j+10

Your answer should be exact (do not round off).

Answer:

The solution is j= and k= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is jk.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by jk, which is the LCD of the fractions. That will cancel all of the denominators.

2j=2k+10jkjk(2j)=jk(2k+10jk)2jjk=2kjk+102k=2j+10

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

2k=2j+104k=3j+10

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by 2 so that the k-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

2(2k)=2(2j+10)4k=4j+20

Now do the elimination and complete the solution to find j.

4k=4j+20second equationSubtract the(4k)=(3j+10)4k+4k=10+204j+3j0=j+10j=10

Great - we have the first value, j=10.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of k. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

4k=3j+104k=3(10)+104k=20k=5

The answer is the pair of numbers j=10 and k=5. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force j,k0, which is why there are open intervals at the j- and k-intercepts of the fraction equation.

The values which solve these two equations are j=10 and k=5.


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ID is: 3289 Seed is: 8821

Equations with fractions

Determine the values of x and y which solve these two equations:

4x3y=133x+7xy=1yx,y0

Your answer should be exact (do not round off).

Answer:

The solution is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is xy.


STEP: Rearrange the second equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by xy, which is the LCD of the fractions. That will cancel all of the denominators.

3x+7xy=1yxy(3x+7xy)=xy(1y)3xxy+7=1yxy3y+7=x

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

4x3y=133y+7=x

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations invite elimination, because the y-terms have equal and opposite coefficients. We can eliminate the y-terms by adding the equations.

Eliminate the y and solve for x:

4x3y=13second equationAdd the+(3y+7)=+(x)74x3y+3y=13x4x+7=x+133x=6x=2

Great - we have the first value, x=2.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of y. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

4x3y=134(2)3y=133y=5y=53

The answer is the pair of numbers x=2 and y=53. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force x,y0, which is why there are open intervals at the x- and y-intercepts of the fraction equation.

The values which solve these two equations are x=2 and y=53.


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ID is: 3289 Seed is: 3261

Equations with fractions

What are the values of a and b which solve these equations simultaneously?

6ab=95b+3a=12aba,b0

Your answer should be exact (do not round off).

Answer:

The solution is a= and b= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is ab.


STEP: Rearrange the second equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by ab, which is the LCD of the fractions. That will cancel all of the denominators.

5b+3a=12abab(5b+3a)=ab(12ab)3aab+5bab=125a+3b=12

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

6ab=95a+3b=12

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by 3 so that the b-coefficients will be ready to cancel. Then we can eliminate them by adding the equations.

First modify things to set up the elimination:

3(6ab)=3(9)18a3b=27

Now do the elimination and complete the solution to find a.

18a3b=27second equationAdd the+(5a+3b)=+(12)18a+5a3b+3b=12+2713a=39a=3

Great - we have the first value, a=3.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of b. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

6ab=96(3)b=9b=9b=9

The answer is the pair of numbers a=3 and b=9. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force a,b0, which is why there are open intervals at the a- and b-intercepts of the fraction equation.

The values which solve these two equations are a=3 and b=9.


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ID is: 395 Seed is: 5725

Solving simultaneous equations by elimination

Solve for x and y using the elimination method:

2x+y=184x+3y=16
Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the y-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Start with the y-terms
[−2 points ⇒ 4 / 6 points left]

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second y-term has a coefficient which is a multiple of the first y-term. In fact, 3 is 3 times as much as 1. We can make both coefficients of y the same by using this factor of 3 to multiply the entire first equation:

2x+y=18(3)(2x)+(3)y=18(3)6x+3y=54

STEP: Eliminate the y-variable and solve for x
[−2 points ⇒ 2 / 6 points left]

Now we can eliminate the y-terms by subtracting one equation from the other. Then we can solve for x.

6x+3y=54to cancel the y-terms:Subtract the equations(4x+3y)=(16)10x+0y=70

Now we can solve for x:

x=7010=7

STEP: Substitute in for x
[−2 points ⇒ 0 / 6 points left]

The last step is to substitute y back into either of the equations so that we can find x. Here we will use the first equation:

2x+y=182(7)+y=1814+y=18y=4

The answers are x=7 and y=4.


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ID is: 395 Seed is: 5686

Solving simultaneous equations by elimination

Solve for x and y using the elimination method:

x+3y=136x+5y=13
Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the x-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Start with the x-terms
[−2 points ⇒ 4 / 6 points left]

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second x-term has a coefficient which is a multiple of the first x-term. In fact, 6 is 6 times as much as 1. We can make both coefficients of x the same by using this factor of 6 to multiply the entire first equation:

x+3y=13(6)x+(6)3y=13(6)6x+18y=78

STEP: Eliminate the x-variable and solve for y
[−2 points ⇒ 2 / 6 points left]

Now we can eliminate the x-terms by subtracting one equation from the other. Then we can solve for y.

6x+18y=78to cancel the x-terms:Subtract the equations(6x+5y)=(13)0x+13y=91

Now we can solve for y:

y=9113=7

STEP: Substitute in for y
[−2 points ⇒ 0 / 6 points left]

The last step is to substitute y back into either of the equations so that we can find x. Here we will use the first equation:

x+3y=13x+3(7)=13x+21=13x=8

The answers are x=8 and y=7.


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ID is: 395 Seed is: 9747

Solving simultaneous equations by elimination

Solve for x and y using the elimination method:

3x+5y=192x+10y=14
Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]
Look at the coefficients of the y-terms. They are not equal, but one of them is a multiple of the other; that is the key to using the elimination method to solve simultaneous equations.
STEP: Start with the y-terms
[−2 points ⇒ 4 / 6 points left]

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second y-term has a coefficient which is a multiple of the first y-term. In fact, 10 is 2 times as much as 5. We can make both coefficients of y the same by using this factor of 2 to multiply the entire first equation:

3x+5y=19(2)3x+(2)5y=19(2)6x+10y=38

STEP: Eliminate the y-variable and solve for x
[−2 points ⇒ 2 / 6 points left]

Now we can eliminate the y-terms by subtracting one equation from the other. Then we can solve for x.

6x+10y=38to cancel the y-terms:Subtract the equations(2x+10y)=(14)8x+0y=24

Now we can solve for x:

x=248=3

STEP: Substitute in for x
[−2 points ⇒ 0 / 6 points left]

The last step is to substitute y back into either of the equations so that we can find x. Here we will use the first equation:

3x+5y=193(3)+5y=199+5y=195y=10y=2

The answers are x=3 and y=2.


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ID is: 3269 Seed is: 5478

Simultaneous equations: the substitution method

Here are two equations, which we can solve simultaneously using substitution:

y=x+55x=15+5y

If we solve the equations using substitution, which variable is the best choice for the substitution, and why? Choose your answers from the choices below.

Answer: The best choice is to substitute because . 2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.


STEP: Look at the arrangement of the equations
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two x-terms and two y-terms:

y=x+55x=15+5y

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the y from the first equation into the second equation. This is because in the first equation the y is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

5x=15+5yy=x+5

Combining these equations by substitution, we get:

5x=15+5(x+5)

Since the first equation tells us that y is equal to y=x+5, we can substitute it into the other equation in place of y straight away.

The best choice is to substitute the y from the first equation into the second equation because it is already isolated.


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ID is: 3269 Seed is: 3530

Simultaneous equations: the substitution method

Here are two equations, which we can solve simultaneously using substitution:

4x=62yx=y3

If we solve the equations using substitution, which variable is the best choice for the substitution, and why? Choose your answers from the choices below.

Answer: The best choice is to substitute because . 2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.


STEP: Look at the arrangement of the equations
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two x-terms and two y-terms:

4x=62yx=y3

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the x from the second equation into the first equation. This is because in the second equation the x is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

4x=62yx=y3

Combining these equations by substitution, we get:

4(y3)=62y

Since the second equation tells us that x is equal to x=y3, we can substitute it into the other equation in place of x straight away.

The best choice is to substitute the x from the second equation into the first equation because it is already isolated.


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ID is: 3269 Seed is: 2708

Simultaneous equations: the substitution method

Here are two equations, which we can solve simultaneously using substitution:

2x=22yx=4y+4

If we solve the equations using substitution, which variable is the best choice for the substitution, and why? Choose your answers from the choices below.

Answer: The best choice is to substitute because . 2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.


STEP: Look at the arrangement of the equations
[−2 points ⇒ 0 / 2 points left]

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two x-terms and two y-terms:

2x=22yx=4y+4

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the x from the second equation into the first equation. This is because in the second equation the x is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

2x=22yx=4y+4

Combining these equations by substitution, we get:

2(4y+4)=22y

Since the second equation tells us that x is equal to x=4y+4, we can substitute it into the other equation in place of x straight away.

The best choice is to substitute the x from the second equation into the first equation because it is already isolated.


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ID is: 3277 Seed is: 8733

Setting up elimination: picking a multiple

A friend in your maths class started to solve these two equations simultaneously:

10=5m4n1=4m+5n

She is using the elimination method. She started by multiplying the second equation by 4, leading to this:

10=5m4n4=16m20n

But your friend is not sure what to do next, and she asks you for help. What number can you use to multiply the first equation so that the equations can be solved using elimination? Your answer should be an integer. Note that there may be more than one answer, but you should only give one answer.

Answer: You can multiply by .
one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the goal of elimination is to cancel either the m-terms or the n-terms. So you need to change one of the coefficients in the first equations so that you can cancel terms.


STEP: Find a way to make two of the terms cancel
[−1 point ⇒ 0 / 1 points left]

For this question we need to multiply the first equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The second equation was already multiplied by 4. We need to change one of the coefficients in the first equation so that we can cancel terms.

Specifically, we need to multiply 5 to get 16 so we can cancel the m-terms, or multiply the 4 to get 20 so we can cancel the n-terms.

The numbers in these equations point us toward cancelling the n-terms. This is because 20 is a multiple of 4. So if we multiply the first equation by 5 the the coefficients will be equal.

term in the equationmultiply each10(5)=5(5)m4(5)n50=25m20n

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

50=25m20nequationsubtract this4=16m20n50+4=25m+16m20n+20n54=9m+0n
NOTE: If we had multiplied by 5, the signs in the first equation would be changed. Then we would add the equations instead. That means that 5 is also an acceptable answer.

The correct answer can be either of these numbers: 5 or 5.


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ID is: 3277 Seed is: 2434

Setting up elimination: picking a multiple

A friend in your maths class started to solve these two equations simultaneously:

5x+4y=104x+3y=7

He is using the elimination method. He started by multiplying the first equation by 4, leading to this:

20x+16y=404x+3y=7

But your friend is not sure what to do next, and he asks you for help. What number can you use to multiply the second equation so that the equations can be solved using elimination? Your answer should be an integer. Note that there may be more than one answer, but you should only give one answer.

Answer: You can multiply by .
one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the goal of elimination is to cancel either the x-terms or the y-terms. So you need to change one of the coefficients in the second equations so that you can cancel terms.


STEP: Find a way to make two of the terms cancel
[−1 point ⇒ 0 / 1 points left]

For this question we need to multiply the second equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The first equation was already multiplied by 4. We need to change one of the coefficients in the second equation so that we can cancel terms.

Specifically, we need to multiply 4 to get 20 so we can cancel the x-terms, or multiply the 3 to get 16 so we can cancel the y-terms.

The numbers in these equations point us toward cancelling the x-terms. This is because 20 is a multiple of 4. So if we multiply the second equation by 5 the the coefficients will be equal.

term in the equationmultiply each4(5)x+3(5)y=7(5)20x+15y=35

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

20x+16y=40equationsubtract this20x+15y=3520x20x+16y15y=40350x+y=5
NOTE: If we had multiplied by 5, the signs in the second equation would be changed. Then we would add the equations instead. That means that 5 is also an acceptable answer.

The correct answer can be either of these numbers: 5 or 5.


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ID is: 3277 Seed is: 8207

Setting up elimination: picking a multiple

A friend in your maths class started to solve these two equations simultaneously:

5k2p=73k3p=3

He is using the elimination method. He started by multiplying the second equation by 2, leading to this:

5k2p=76k+6p=6

But your friend is not sure what to do next, and he asks you for help. What number can you use to multiply the first equation so that the equations can be solved using elimination? Your answer should be an integer. Note that there may be more than one answer, but you should only give one answer.

Answer: You can multiply by .
one-of
type(numeric.noerror)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Remember that the goal of elimination is to cancel either the k-terms or the p-terms. So you need to change one of the coefficients in the first equations so that you can cancel terms.


STEP: Find a way to make two of the terms cancel
[−1 point ⇒ 0 / 1 points left]

For this question we need to multiply the first equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The second equation was already multiplied by 2. We need to change one of the coefficients in the first equation so that we can cancel terms.

Specifically, we need to multiply 5 to get 6 so we can cancel the k-terms, or multiply the 2 to get 6 so we can cancel the p-terms.

The numbers in these equations point us toward cancelling the p-terms. This is because 6 is a multiple of 2. So if we multiply the first equation by 3 the the coefficients will be equal and opposite.

term in the equationmultiply each5(3)k2(3)p=7(3)15k6p=21

The coefficients we want to cancel are equal and opposite. So eliminating the terms requires adding both sides of the equation:

15k6p=21equationadd this6k+6p=615k6k6p+6p=21+69k+0p=27
NOTE: If we had multiplied by 3, the signs in the first equation would be changed. Then we would subtract the equations instead. That means that 3 is also an acceptable answer.

The correct answer can be either of these numbers: 3 or 3.


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ID is: 3268 Seed is: 4730

Picking a solution method

The following equations both include the variables x and y:

y=3x55x=102y
  1. These equations can be solved using either the elimination or the substitution method. Which method is better for solving these two equations?

    Answer: The better method is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To decide which method is better, you should look at the coefficients. For example, can the coefficients lead to the cancellation of any of the terms in these equations?


    STEP: Check the arrangement of the equations to decide which method is better
    [−1 point ⇒ 0 / 1 points left]

    This question shows us two equations. We need to decide which method is the best choice for solving the equations simultaneously. (We do not have to solve the equations!)

    Two methods we can use for solving equations simultaneously are elimination and substitution. Both of those methods can be used to solve the equations in the question. But usually one of the methods is easier than the other based on the equations. Deciding which method is better depends on the arrangement of the equations:

    • Substitution is the better choice if there is already one variable isolated or if one variable is easy to isolate. If a variable is isolated, we can substitute immediately.
    • Elimination is the better choice if there are two terms which can be cancelled easily. If the terms can cancel, we can eliminate those terms immediately.

    The equations here are:

    y=3x55x=102y

    In this case, the first equation points us to substitution: the y in that equation is isolated. This is the perfect arrangement for substitution, because we can substitute y=3x5 directly into the y in the second equation.

    The correct answer is: the substitution method.


    Submit your answer as:
  2. Solve the equations simultaneously.

    INSTRUCTION: Type your answer as a coordinate pair like this: (x;y).
    Answer: The answer is: .
    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the method identified in Question 1.


    STEP: Combine the equations using substitution
    [−1 point ⇒ 3 / 4 points left]

    Based on the result of Question 1, we should solve this question using substitution. It is possible to solve the equations using elimination, but using substitution is easier (as described in Question 1). Substitute y=3x5 into the second equation in place of y:

    5x=102yy=3x5

    Combining these equations by substution, we get:

    5x=102(3x5)

    STEP: Solve for x
    [−2 points ⇒ 1 / 4 points left]

    Now we can solve the equation. Distribute the 2 and get on with solving the equation.

    5x=102(3x5)5x=102(3x)2(5)5x=106x+105x+6x=10+10x=0

    STEP: Find the value of y
    [−1 point ⇒ 0 / 4 points left]

    So x=0. But remember that we also need an answer for y. We can find this using the answer we just got for x. It is important to remember that you can use either equation to calculate y, because the numbers we want solve both equations. It is good to pick the easiest choice. In this case, the easier choice is the first equation, because it is arranged in a more useful way.

    y=3x5y=3(0)5y=05y=5

    The numbers which solve the equations simultaneously are x=0 and y=5.

    The correct answer is (0;5).


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ID is: 3268 Seed is: 1354

Picking a solution method

These two equations contain the same variables:

1=3x4y5=x+4y
  1. These equations can be solved using either the elimination or the substitution method. Which method is better for solving these two equations?

    Answer: The better method is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To decide which method is better, you should look at the coefficients. For example, can the coefficients lead to the cancellation of any of the terms in these equations?


    STEP: Check the arrangement of the equations to decide which method is better
    [−1 point ⇒ 0 / 1 points left]

    This question shows us two equations. We need to decide which method is the best choice for solving the equations simultaneously. (We do not have to solve the equations!)

    Two methods we can use for solving equations simultaneously are elimination and substitution. Both of those methods can be used to solve the equations in the question. But usually one of the methods is easier than the other based on the equations. Deciding which method is better depends on the arrangement of the equations:

    • Substitution is the better choice if there is already one variable isolated or if one variable is easy to isolate. If a variable is isolated, we can substitute immediately.
    • Elimination is the better choice if there are two terms which can be cancelled easily. If the terms can cancel, we can eliminate those terms immediately.

    The equations here are:

    1=3x4y5=x+4y

    In this case, the y-terms are ready to be eliminated. The coefficients of these terms are equal and opposite, which is perfect for the elimination method: it means we can cancel those two terms if we add the equations, which is how elimination works.

    The correct answer is: the elimination method.


    Submit your answer as:
  2. Now find the simultaneous solution to the equations.

    INSTRUCTION: Type your answer as a coordinate pair like this: (x;y).
    Answer: The answer is: .
    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Use the method identified in Question 1.


    STEP: Eliminate (cancel) the y-terms
    [−1 point ⇒ 2 / 3 points left]

    Based on the result of Question 1, we should solve this question using elimination. It is possible to solve the equations using substitution, but using elimination is easier (as described in Question 1). Add the equations together to eliminate the y-terms. This means we add together the left and right sides of the equations.

    1=3x4yequationadd this5=x+4y15=3x+x4y+4y6=2x+0y6=2x


    STEP: Solve for x
    [−1 point ⇒ 1 / 3 points left]

    Now we have an equation with only one variable, and we can solve it.

    6=2x3=x

    STEP: Find the value of y
    [−1 point ⇒ 0 / 3 points left]

    So x=3. But we also need an answer for y. We can get this using the answer we just got for x: substitute this into either of the equations to get the answer. We can use either equation because we are looking for the number that solves both of them. Since we have a choice, we should pick the easier equation (if there is one). We will pick the second equation, because it has fewer negative signs.

    5=x+4y5=(3)+4y5=3+4y8=4y2=y

    Now we have the complete answer: the numbers which solve the equations simultaneously are x=3 and y=2.

    The correct answer is (3;2).


    Submit your answer as:

ID is: 3268 Seed is: 8075

Picking a solution method

Here are two equations, which contain the same variables:

y=2x1x=83y
  1. These equations can be solved using either the elimination or the substitution method. Which method is better for solving these two equations?

    Answer: The better method is .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    To decide which method is better, you should look at the coefficients. For example, can the coefficients lead to the cancellation of any of the terms in these equations?


    STEP: Check the arrangement of the equations to decide which method is better
    [−1 point ⇒ 0 / 1 points left]

    This question shows us two equations. We need to decide which method is the best choice for solving the equations simultaneously. (We do not have to solve the equations!)

    Two methods we can use for solving equations simultaneously are elimination and substitution. Both of those methods can be used to solve the equations in the question. But usually one of the methods is easier than the other based on the equations. Deciding which method is better depends on the arrangement of the equations:

    • Substitution is the better choice if there is already one variable isolated or if one variable is easy to isolate. If a variable is isolated, we can substitute immediately.
    • Elimination is the better choice if there are two terms which can be cancelled easily. If the terms can cancel, we can eliminate those terms immediately.

    The equations here are:

    y=2x1x=83y

    In this case, the first equation points us to substitution: the y in that equation is isolated. This is the perfect arrangement for substitution, because we can substitute y=2x1 directly into the y in the second equation.

    The correct answer is: the substitution method.


    Submit your answer as:
  2. Now find the simultaneous solution to the equations.

    INSTRUCTION: Type your answer as a coordinate pair like this: (x;y).
    Answer: The answer is: .
    coordinate
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the method identified in Question 1.


    STEP: Combine the equations using substitution
    [−1 point ⇒ 3 / 4 points left]

    Based on the result of Question 1, we should solve this question using substitution. It is possible to solve the equations using elimination, but using substitution is easier (as described in Question 1). Substitute y=2x1 into the second equation in place of y:

    x=83yy=2x1

    Combining these equations by substution, we get:

    x=83(2x1)

    STEP: Solve for x
    [−2 points ⇒ 1 / 4 points left]

    Now we can solve the equation. Distribute the 3 and get on with solving the equation.

    x=83(2x1)x=83(2x)3(1)x=86x+3x+6x=8+35x=5x=1

    STEP: Find the value of y
    [−1 point ⇒ 0 / 4 points left]

    So x=1. But remember that we also need an answer for y. We can find this using the answer we just got for x. It is important to remember that you can use either equation to calculate y, because the numbers we want solve both equations. It is good to pick the easiest choice. In this case, the easier choice is the first equation, because it is arranged in a more useful way.

    y=2x1y=2(1)1y=21y=3

    The numbers which solve the equations simultaneously are x=1 and y=3.

    The correct answer is (1;3).


    Submit your answer as:

ID is: 3266 Seed is: 8891

Simultaneous equations on the Cartesian plane

The graph below shows these two equations:

y=x1y=3x+1

What is the solution to these two equations when they are solved simultaneously? Use the graph to answer the question.

Answer:

The solution to the equations is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to find the coordinates of the point where the lines intersect.


STEP: Read the answer from the point of intersection
[−2 points ⇒ 0 / 2 points left]

This question is about two equations. We need to find the values of x and y which solve the equations 'simultaneously'. This means we want one pair of x and y values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point (1;2).

We can prove that this is correct by substituting the values into each of the equations:

y=x1(2)=(1)12=2

and

y=3x+1(2)=3(1)+12=2

Perfect - the values x=1 and y=2 solve the equations simultaneously.

The correct answers are x=1 and y=2.


Submit your answer as: and

ID is: 3266 Seed is: 523

Simultaneous equations on the Cartesian plane

The graph below shows these two equations:

y=x3+3y=2x2

What is the solution to these two equations when they are solved simultaneously? Use the graph to answer the question.

Answer:

The solution to the equations is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to find the coordinates of the point where the lines intersect.


STEP: Read the answer from the point of intersection
[−2 points ⇒ 0 / 2 points left]

This question is about two equations. We need to find the values of x and y which solve the equations 'simultaneously'. This means we want one pair of x and y values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point (3;4).

We can prove that this is correct by substituting the values into each of the equations:

y=x3+3(4)=(3)3+34=4

and

y=2x2(4)=2(3)24=4

Perfect - the values x=3 and y=4 solve the equations simultaneously.

The correct answers are x=3 and y=4.


Submit your answer as: and

ID is: 3266 Seed is: 5914

Simultaneous equations on the Cartesian plane

The graph below shows these two equations:

y=x1y=x2+2

What is the solution to these two equations when they are solved simultaneously? Use the graph to answer the question.

Answer:

The solution to the equations is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to find the coordinates of the point where the lines intersect.


STEP: Read the answer from the point of intersection
[−2 points ⇒ 0 / 2 points left]

This question is about two equations. We need to find the values of x and y which solve the equations 'simultaneously'. This means we want one pair of x and y values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point (2;1).

We can prove that this is correct by substituting the values into each of the equations:

y=x1(1)=(2)11=1

and

y=x2+2(1)=(2)2+21=1

Perfect - the values x=2 and y=1 solve the equations simultaneously.

The correct answers are x=2 and y=1.


Submit your answer as: and

ID is: 368 Seed is: 9299

Solving simultaneous equations by substitution

Solve simultaneously for x and y.

3x+y=112x+6y=14
TIP: Use the substitution method.
Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.


STEP: Identify which variable to isolate
[−1 point ⇒ 5 / 6 points left]

When we solve simultaneous equations, we are working to find values for y and x which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the first equation looks good for this, because the y has a coefficient of one:

3x+y=11 isolationtarget for

STEP: Make y the subject of the equation
[−1 point ⇒ 4 / 6 points left]

Isolate y in the first equation, which can be done in one step:

3x+y=11y=3x+11

STEP: Substitute the result into the other equation to find x
[−2 points ⇒ 2 / 6 points left]

Now we can substitute the expression 3x+11 into the second equation and solve:

2x+6y=142x+6(3x+11)=142x+18x+66=1416x=80x=5

STEP: Use the x-value to find y
[−2 points ⇒ 0 / 6 points left]

Right now we have only half of the answer! To get the other half, we need to find the value of y. We can use the equation we got when we isolated y: it is the most convenient choice because y is the subject of the equation.

y=3x+11y=3(5)+11y=4

The correct answers are x=5 and y=4.


Submit your answer as: and

ID is: 368 Seed is: 2411

Solving simultaneous equations by substitution

Solve simultaneously for x and y.

x4y=264x6y=64
TIP: Use the substitution method.
Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.


STEP: Identify which variable to isolate
[−1 point ⇒ 5 / 6 points left]

When we solve simultaneous equations, we are working to find values for x and y which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the first equation looks good for this, because the x has a coefficient of one:

x4y=26 isolationtarget for

STEP: Make x the subject of the equation
[−1 point ⇒ 4 / 6 points left]

Isolate x in the first equation, which can be done in one step:

x4y=26x=4y+26

STEP: Substitute the result into the other equation to find y
[−2 points ⇒ 2 / 6 points left]

Now we can substitute the expression 4y+26 into the second equation and solve:

4x6y=644(4y+26)6y=6416y+1046y=6410y=40y=4

STEP: Use the y-value to find x
[−2 points ⇒ 0 / 6 points left]

Right now we have only half of the answer! To get the other half, we need to find the value of x. We can use the equation we got when we isolated x: it is the most convenient choice because x is the subject of the equation.

x=4y+26x=4(4)+26x=10

The correct answers are x=10 and y=4.


Submit your answer as: and

ID is: 368 Seed is: 1423

Solving simultaneous equations by substitution

Solve simultaneously for x and y.

5x+4y=02x+y=3
TIP: Use the substitution method.
Answer: x= and y=
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.


STEP: Identify which variable to isolate
[−1 point ⇒ 5 / 6 points left]

When we solve simultaneous equations, we are working to find values for y and x which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the y has a coefficient of one:

2x+y=3 isolationtarget for

STEP: Make y the subject of the equation
[−1 point ⇒ 4 / 6 points left]

Isolate y in the second equation, which can be done in one step:

2x+y=3y=2x3

STEP: Substitute the result into the other equation to find x
[−2 points ⇒ 2 / 6 points left]

Now we can substitute the expression 2x3 into the first equation and solve:

5x+4y=05x+4(2x3)=05x+8x12=03x=12x=4

STEP: Use the x-value to find y
[−2 points ⇒ 0 / 6 points left]

Right now we have only half of the answer! To get the other half, we need to find the value of y. We can use the equation we got when we isolated y: it is the most convenient choice because y is the subject of the equation.

y=2x3y=2(4)3y=5

The correct answers are x=4 and y=5.


Submit your answer as: and

2. Linear & quadratic equations


ID is: 3306 Seed is: 7820

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+6x+5y=2x10
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+6x+5 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+6x+5.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is 5. The coordinates of this point are (0;5).

The x-intercepts are the roots of the equation x2+6x+5=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 5. This is the product of the coefficient of the first term 1 and the value of the last term 5. The two numbers are 1 and 5. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+6x+50=x2+x+5x+5

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+1)+5(x+1)0=(x+5)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1 and x=5. The coordinates of these roots are (1;0) and (5;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=2x10 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 10. Its coordinates are: (0;10).

We will work out the root of the equation 0=2x10.

0=2x10=02x=10x=5

The root is: x=5. It's coordinates are: (5;0).

We will now plot the equation y=2x10 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (3;4) and (5;0).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+6x+5 and y=2x10. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (3;4) and (5;0).


Submit your answer as: and

ID is: 3306 Seed is: 320

Points of intersection

Sketch the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x27x+6y=2x+2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x27x+6 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x27x+6.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is 6. The coordinates of this point are (0;6).

The x-intercepts are the roots of the equation x27x+6=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 7. This is the coefficient of the middle term. The product of the two numbers must be 6. This is the product of the coefficient of the first term 1 and the value of the last term 6. The two numbers are 1 and 6. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x27x+60=x2x6x+6

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x1)6(x1)0=(x6)(x1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=6 and x=1. The coordinates of these roots are (6;0) and (1;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=2x+2 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 2. Its coordinates are: (0;2).

We will work out the root of the equation 0=2x+2.

0=2x+2=02x=2x=1

The root is: x=1. It's coordinates are: (1;0).

We will now plot the equation y=2x+2 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (4;6) and (1;0).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x27x+6 and y=2x+2. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (4;6) and (1;0).


Submit your answer as: and

ID is: 3306 Seed is: 7244

Points of intersection

Graph the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+3x4y=x+1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+3x4 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+3x4.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is −4. The coordinates of this point are (0;4).

The x-intercepts are the roots of the equation x2+3x4=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 3. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. The two numbers are 4 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+3x40=x2+4xx4

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+4)(x+4)0=(x1)(x+4)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1 and x=4. The coordinates of these roots are (1;0) and (4;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=x+1 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 1. Its coordinates are: (0;1).

We will work out the root of the equation 0=x+1.

0=x+1=0x=1x=1

The root is: x=1. It's coordinates are: (1;0).

We will now plot the equation y=x+1 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;0) and (5;6).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+3x4 and y=x+1. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;0) and (5;6).


Submit your answer as: and

ID is: 1401 Seed is: 2107

Simultaneous equations: solving by substitution

Find the complete solution for these simultaneous equations. You should get two coordinate pairs, (x;y), for your answers.

y+4=x+164=x2+5xy
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate y.

y+4=x+16y=x+12

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (x+12), into the quadratic equation in place of y. Be sure to use brackets! Then simplify the equation to get standard form.

4=x2+5xy4=x2+5x(x+12)4=x2+6x12x26x+8=0to keep the quadratic term positivemove the terms onto the left side

STEP: Solve for the values of x
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two x-values.

x26x+8=0(x4)(x2)=0x=4 and x=2

STEP: Find the value of y for each of the x-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of y for these equations. We must substitute in the values x=4 and x=2. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=4:y=(4)+12y=8If x=2:y=(2)+12y=10

Finally we have the answers: if x=4 then y=8 and if x=2 then y=10.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (4;8) and (2;10).


Submit your answer as: and

ID is: 1401 Seed is: 8275

Simultaneous equations: solving by substitution

Solve for x and y. You should get two coordinate pairs, (x;y), for your answers.

10=x+3yx8=y2+6y
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

10=x+3yx=3y+10

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (3y+10), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

x8=y2+6y(3y+10)8=y2+6y3y+2=y2+6yy23y+2=0to keep the quadratic term positivemove the terms onto the left side

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

y23y+2=0(y2)(y1)=0y=1 and y=2

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=1 and y=2. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=1:x=3(1)+10x=13If y=2:x=3(2)+10x=16

Finally we have the answers: if x=13 then y=1 and if x=16 then y=2.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (13;1) and (16;2).


Submit your answer as: and

ID is: 1401 Seed is: 4882

Simultaneous equations: solving by substitution

Solve this pair of simultaneous equations. You should get two coordinate pairs, (x;y), for your answers.

17=x6yy2+6y=x+8
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

17=x6yx=6y+17

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (6y+17), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

y2+6y=x+8y2+6y=(6y+17)+8y2+6y=6y90=y29to keep the quadratic term positivemove the terms onto the right side

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

0=y290=(y3)(y+3)y=3 and y=3

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=3 and y=3. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=3:x=6(3)+17x=35If y=3:x=6(3)+17x=1

Finally we have the answers: if x=35 then y=3 and if x=1 then y=3.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (35;3) and (1;3).


Submit your answer as: and

ID is: 3307 Seed is: 2042

Simultaneous equations

Solve for x and y in the following simultaneous equations:

y=2x26x+2y=2x4
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x4 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x26x+2linear equationy=2x4

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

2x4=2x26x+20=2x26x+22x+40=2x26x2x+2+40=2x28x+6

STEP: Solve the equation 2x28x+6=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 8. This is the coefficient of the middle term. The product of the two numbers must be 12. This is the product of the coefficient of the first term 2 and the value of the last term 6. The two numbers are 2 and 6. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x28x+60=2x22x6x+6

We now group the first two terms together and group the last terms together and factorise.

0=2x22x6x+60=2x(x1)6(x1)

The last thing now is to factor the highest common factor.

0=(2x6)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 3 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=3:

y=2x4=2

For x=1:

y=2x4=2

The solutions are x=3 together with y=2 , and x=1 with y=2. In coordinates pairs, the solutions are (3;2) and (1;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;2) and (1;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;2) and (1;2).


Submit your answer as: and

ID is: 3307 Seed is: 9542

Simultaneous equations

Solve for x and y:

y=x24x+1y=x3
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x24x+1linear equationy=x3

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

x3=x24x+10=x24x+1x+30=x24xx+1+30=x25x+4

STEP: Solve the equation x25x+4=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. The two numbers are 1 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x25x+40=x2x4x+4

We now group the first two terms together and group the last terms together and factorise.

0=x2x4x+40=x(x1)4(x1)

The last thing now is to factor the highest common factor.

0=(x4)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 4 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=4:

y=x3=1

For x=1:

y=x3=2

The solutions are x=4 together with y=1 , and x=1 with y=2. In coordinates pairs, the solutions are (4;1) and (1;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;1) and (1;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;1) and (1;2).


Submit your answer as: and

ID is: 3307 Seed is: 6281

Simultaneous equations

Solve for x and y:

y=2x29x+8y=3x8
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 3x8 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x29x+8linear equationy=3x8

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

3x8=2x29x+80=2x29x+83x+80=2x29x3x+8+80=2x212x+16

STEP: Solve the equation 2x212x+16=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 12. This is the coefficient of the middle term. The product of the two numbers must be 32. This is the product of the coefficient of the first term 2 and the value of the last term 16. The two numbers are 4 and 8. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x212x+160=2x24x8x+16

We now group the first two terms together and group the last terms together and factorise.

0=2x24x8x+160=2x(x+2)+8(x+2)

The last thing now is to factor the highest common factor.

0=(2x+8)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 4 and 2.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=4:

y=3x8=4

For x=2:

y=3x8=2

The solutions are x=4 together with y=4 , and x=2 with y=2. In coordinates pairs, the solutions are (4;4) and (2;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;4) and (2;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;4) and (2;2).


Submit your answer as: and

ID is: 3303 Seed is: 1543

Simultaneous equations

Solve for x and y:

y=3x2+5x1y=x+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x2+5x1linear equationy=x+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x+3=3x2+5x10=3x2+5x1x30=3x2+5xx130=3x2+4x4

STEP: Solve the equation 0=3x2+4x4
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 12. This is the product of the coefficient of the first term 3 and the value of the last term 4. The two numbers are 6 and 2. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x2+4x40=3x2+6x2x4

We now group the first two terms together and group the last terms together and factorise.

0=3x2+6x2x40=3x(x+2)2(x+2)

The last thing now is to factor the highest common factor.

0=(3x2)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 23 and 2.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=23:

y=x+3=113

For x=2:

y=x+3=1

The solutions are x=23 together with y=113 , and x=2 with y=1. In coordinate pairs, the solutions are (23;113) and (2;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (23;113) and (2;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (23;113) and (2;1).


Submit your answer as: and

ID is: 3303 Seed is: 4866

Simultaneous equations

Solve the following equations simultaneously:

y=2x2+2x3y=x1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x2+2x3linear equationy=x1

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x1=2x2+2x30=2x2+2x3+x+10=2x2+2x+x3+10=2x2+3x2

STEP: Solve the equation 0=2x2+3x2
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 3. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 2 and the value of the last term 2. The two numbers are 4 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x2+3x20=2x2+4xx2

We now group the first two terms together and group the last terms together and factorise.

0=2x2+4xx20=2x(x+2)(x+2)

The last thing now is to factor the highest common factor.

0=(2x1)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 12 and 2.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=12:

y=x1=32

For x=2:

y=x1=1

The solutions are x=12 together with y=32 , and x=2 with y=1. In coordinate pairs, the solutions are (12;32) and (2;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (12;32) and (2;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (12;32) and (2;1).


Submit your answer as: and

ID is: 3303 Seed is: 347

Simultaneous equations

Solve the following equations simultaneously:

y=3x2+3x3y=2x1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x2+3x3linear equationy=2x1

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x1=3x2+3x30=3x2+3x3+2x+10=3x2+3x+2x3+10=3x2+5x2

STEP: Solve the equation 0=3x2+5x2
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 6. This is the product of the coefficient of the first term 3 and the value of the last term 2. The two numbers are 6 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x2+5x20=3x2+6xx2

We now group the first two terms together and group the last terms together and factorise.

0=3x2+6xx20=3x(x+2)(x+2)

The last thing now is to factor the highest common factor.

0=(3x1)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 13 and 2.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=13:

y=2x1=53

For x=2:

y=2x1=3

The solutions are x=13 together with y=53 , and x=2 with y=3. In coordinate pairs, the solutions are (13;53) and (2;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (13;53) and (2;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (13;53) and (2;3).


Submit your answer as: and

ID is: 3308 Seed is: 3384

Simultaneous equations

Solve for x and y:

y=x2+3x+4x=y4
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y4 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+3x+4linear equationx=y4

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y4)2+3(y4)+40=(y4)2+3(y4)+4y0=y28y+16+(3y12)+4y0=y28y+16+3y12+4y0=y28y+3yy+1612+40=y25yy+1612+40=y26y+1612+40=y26y+4+40=y26y+8

STEP: Solve the equation 0=y26y+8
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 8. This is the product of the coefficient of the first term 1 and the value of the last term 8. The two numbers are 2 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y26y+80=y22y4y+8

We now group the first two terms together and group the last terms together and factorise.

0=y22y4y+80=y(y+2)+4(y+2)

The last thing now is to factor the highest common factor.

0=(y+4)(y+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 4 and 2.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=4:

x=y4=0

For y=2:

x=y4=2

The solutions are x=0 together with y=4 , and x=2 with y=2. In coordinates pairs, the solutions are (0;4) and (2;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (0;4) and (2;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (0;4) and (2;2).


Submit your answer as: and

ID is: 3308 Seed is: 1630

Simultaneous equations

Solve for x and y:

y=x2+3x+4x=y4
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y4 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+3x+4linear equationx=y4

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y4)2+3(y4)+40=(y4)2+3(y4)+4y0=y28y+16+(3y12)+4y0=y28y+16+3y12+4y0=y28y+3yy+1612+40=y25yy+1612+40=y26y+1612+40=y26y+4+40=y26y+8

STEP: Solve the equation 0=y26y+8
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 8. This is the product of the coefficient of the first term 1 and the value of the last term 8. The two numbers are 2 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y26y+80=y22y4y+8

We now group the first two terms together and group the last terms together and factorise.

0=y22y4y+80=y(y+2)+4(y+2)

The last thing now is to factor the highest common factor.

0=(y+4)(y+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 4 and 2.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=4:

x=y4=0

For y=2:

x=y4=2

The solutions are x=0 together with y=4 , and x=2 with y=2. In coordinates pairs, the solutions are (0;4) and (2;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (0;4) and (2;2). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (0;4) and (2;2).


Submit your answer as: and

ID is: 3308 Seed is: 799

Simultaneous equations

Solve for x and y from the given equations:

y=x23x1x=y+1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+1 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x23x1linear equationx=y+1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y+1)23(y+1)10=(y+1)23(y+1)1y0=y2+2y+1(3y+3)1y0=y2+2y+13y31y0=y2+2y3yy+1310=y2yy+1310=y22y+1310=y22y210=y22y3

STEP: Solve the equation 0=y22y3
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 2. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y22y30=y2+y3y3

We now group the first two terms together and group the last terms together and factorise.

0=y2+y3y30=y(y+1)3(y+1)

The last thing now is to factor the highest common factor.

0=(y3)(y+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 3 and -1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=3:

x=y+1=4

For y=1:

x=y+1=0

The solutions are x=4 together with y=3 , and x=0 with y=1. In coordinates pairs, the solutions are (4;3) and (0;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;3) and (0;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;3) and (0;1).


Submit your answer as: and

ID is: 1399 Seed is: 1747

Simultaneous equations: intersection points

The graph below shows the equations:

y=x232andy=x22+x+32

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity x232; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is x232and:yis also x22+x+32Therefore: x232=x22+x+32

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 2. So multiply both sides of the equation by 2 in order to cancel the denominators. Then arrange the equation in standard form.

(2)(x232)=(x22+x+32)(2)x3=x2+2x+30=x2+x+6

Remember that if the quadratic coefficient is negative, we can multiply the equation by −1. That will make it easier to factorise.

0=x2x6

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2x60=(x3)(x+2)x=2 and x=3

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=2 and x=3 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=2:y=(2)232y=52If x=3:y=(3)232y=0

Finally! The solutions are (2;52) and (3;0).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (2;52) and (3;0).


Submit your answer as: and

ID is: 1399 Seed is: 3757

Simultaneous equations: intersection points

The graph below shows the equations:

y=16x3+43andy=4x23+43

Find the intersection points of the curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity 16x3+43; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is 16x3+43and:yis also 4x23+43Therefore: 16x3+43=4x23+43

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(16x3+43)=(4x23+43)(3)16x+4=4x2+40=4x2+16x

In this case, there is a common factor of −4 for all of the terms. Divide out this factor before factorising.

0=x24x

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x24x0=x(x4)x=0 and x=4

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=0 and x=4 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=0:y=16(0)3+43y=43If x=4:y=16(4)3+43y=20

Finally! The solutions are (0;43) and (4;20).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (0;43) and (4;20).


Submit your answer as: and

ID is: 1399 Seed is: 8817

Simultaneous equations: intersection points

The graph below shows the equations:

y=x3+73andy=x23+x323

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity x3+73; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is x3+73and:yis also x23+x323Therefore: x3+73=x23+x323

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(x3+73)=(x23+x323)(3)x+7=x2+x20=x29


STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x290=(x3)(x+3)x=3 and x=3

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=3 and x=3 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=3:y=(3)3+73y=43If x=3:y=(3)3+73y=103

Finally! The solutions are (3;43) and (3;103).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (3;43) and (3;103).


Submit your answer as: and

ID is: 3305 Seed is: 4325

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x2+2x5x=y1

Which of the following equations shows the correct substitution step?

A x=y2+2y5
B y=(y+1)2+2(y+1)5
C y1=y2+2y
D y=(y1)2+2(y1)5
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute y1 in place of x in the quadratic equation y=x2+2x5.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x2+2x5linear equationx=y1

We need to substitute y1 from the linear equation in place of x in the quadratic equation.

y=(y1)2+2(y1)5
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for y. The question does not ask us to solve the simultaneous equations.

The correct substitution step is y=(y1)2+2(y1)5 which is choice D.


Submit your answer as:

ID is: 3305 Seed is: 9182

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x25x7y=3x4

Which of the following equations shows the correct substitution step?

A y=(3x)25(3x)7
B 3x+4=x25x7
C 3x4=x25x7
D 3x=x25x7
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute 3x4 in place of y in the quadratic equation y=x25x7.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x25x7linear equationy=3x4

We need to substitute 3x4 from the linear equation in place of y in the quadratic equation.

3x4=x25x7
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is 3x4=x25x7 which is choice C.


Submit your answer as:

ID is: 3305 Seed is: 1399

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x28x6y=3x10

Which of the following equations shows the correct substitution step?

A 3x10=y28y6
B 3x10=x28x6
C 3y10=x28x6
D y=(3x)28(3x)6
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute 3x10 in place of y in the quadratic equation y=x28x6.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x28x6linear equationy=3x10

We need to substitute 3x10 from the linear equation in place of y in the quadratic equation.

3x10=x28x6
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is 3x10=x28x6 which is choice B.


Submit your answer as:

ID is: 3304 Seed is: 8843

Simultaneous equations: special outcomes

Solve the following equations simultaneously:

y=x2+4x+5y=2x4
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting 2x4 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=x2+4x+5 and the linear equation is y=2x4. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x4=x2+4x+50=x2+4x+5+2x+40=x2+4x+2x+5+40=x2+6x+9

STEP: Solve the equation 0=x2+6x+9
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 9. This is the product of the coefficient of the first term 1 and the value of the last term 9. If we take 3 which is half of the value of 6 (the coefficient of the middle term) and square it, we get the product 9. In this case we will use only one number to factorise, that is, 3. We will re-write the middle term, forming two terms, using this value.

0=x2+6x+90=x2+3x+3x+9

We now group the first two terms together and group the last terms together and factorise.

0=x2+3x+3x+90=x(x+3)+3(x+3)

The last thing now is to factor the highest common factor.

0=(x+3)(x+3)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 3. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=2x4=2(3)4=2

The solution is x=3 together with y=2. Using coordinate pairs, the solution is (3;2).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;2). This is the same solution we have calculated. This means we are 100% correct.

The solution is (3;2).


Submit your answer as:

ID is: 3304 Seed is: 3329

Simultaneous equations: special outcomes

Solve the following equations simultaneously:

y=3x2xy=3x1
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
one-of
type(string.nocase)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting 3x1 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=3x2x and the linear equation is y=3x1. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
3x1=3x2x0=3x2x+3x+10=3x2x+3x+10=3x2+2x+1

STEP: Solve the equation 0=3x2+2x+1
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give 2. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 3 and the value of the last term 1. In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form ax2+bx+c=0:

x=b±(b24ac)2a

Before we can use the quadratic formula, we need to identify the coefficients a, b and c from the equation 3x2+2x+1=0 These are:

a=3,b=2 and c=1.

We will substitute these values into the formula and simplify.

x=(2)±((2)24(3)(1))2(3)=2±(4(12))6=2±(8)6

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that y=3x2x and y=3x1 cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.


Submit your answer as:

ID is: 3304 Seed is: 8342

Simultaneous equations: special outcomes

Solve for x and y from the given equations:

y=x2x+2y=3x+1
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting 3x+1 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=x2x+2 and the linear equation is y=3x+1. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
3x+1=x2x+20=x2x+2+3x10=x2x+3x+210=x2+2x+1

STEP: Solve the equation 0=x2+2x+1
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 2. This is the coefficient of the middle term. The product of the two numbers must be 1. This is the product of the coefficient of the first term 1 and the value of the last term 1. If we take 1 which is half of the value of 2 (the coefficient of the middle term) and square it, we get the product 1. In this case we will use only one number to factorise, that is, 1. We will re-write the middle term, forming two terms, using this value.

0=x2+2x+10=x2+x+x+1

We now group the first two terms together and group the last terms together and factorise.

0=x2+x+x+10=x(x+1)+(x+1)

The last thing now is to factor the highest common factor.

0=(x+1)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 1. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=3x+1=3(1)+1=4

The solution is x=1 together with y=4. Using coordinate pairs, the solution is (1;4).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;4). This is the same solution we have calculated. This means we are 100% correct.

The solution is (1;4).


Submit your answer as:

ID is: 3852 Seed is: 4237

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=5yandx2+3xy=40
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=5y into x2+3xy=40 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=5yandx2+3xy=40

First, we can substitute 5y in place of x in the second equation:

(5y)2+3(5y)(y)=40

We can simplify this equation to solve for y:

(5y)2+3(5y)(y)=4025y215y2=4010y2=40y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=5y.

For y=2:

x=5y=5(2)=10

And for y=2:

x=5y=5(2)=10

So the x-values are x=10 or x=10.

Therefore the correct coordinate pairs are (10;2) and (10;2).


Submit your answer as: and

ID is: 3852 Seed is: 8256

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=5yandx24xy=45
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=5y into x24xy=45 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=5yandx24xy=45

First, we can substitute 5y in place of x in the second equation:

(5y)24(5y)(y)=45

We can simplify this equation to solve for y:

(5y)24(5y)(y)=4525y220y2=455y2=45y2=9

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=3 or y=3

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=5y.

For y=3:

x=5y=5(3)=15

And for y=3:

x=5y=5(3)=15

So the x-values are x=15 or x=15.

Therefore the correct coordinate pairs are (15;3) and (15;3).


Submit your answer as: and

ID is: 3852 Seed is: 8460

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx25xy=24
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x25xy=24 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx25xy=24

First, we can substitute 2y in place of x in the second equation:

(2y)25(2y)(y)=24

We can simplify this equation to solve for y:

(2y)25(2y)(y)=244y210y2=246y2=24y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=2:

x=2y=2(2)=4

And for y=2:

x=2y=2(2)=4

So the x-values are x=4 or x=4.

Therefore the correct coordinate pairs are (4;2) and (4;2).


Submit your answer as: and

ID is: 3302 Seed is: 7408

Simultaneous equations with xy terms

Solve for x and y:

2y=6xxy+5y=x2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x2 in place of y in the hyperbolic equation and then solve for x.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=6xxy+5linear equationy=x2

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 2y=6xxy+5 and substitute its value into y=x2. We could still get the same answer, but the calculations might become too long.
2(x2)=6xx(x2)+50=6xx(x2)+52(x2)0=6x(x22x)+5(2x4)0=x2+8x+2x+5+40=x2+10x+5+40=x2+10x+9

STEP: Solve the equation 0=x2+10x+9
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+10x+90=(x+1)(x+9)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 1 and 9.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

By using the values of x, we will calculate the corresponding values of y. We will do so by substituting each x-value into the linear equation.

NOTE: We can still get the same values of y if we substitute the values x into the equation 2y=6xxy+5. If we do that, the calculations might become too long.

For x=1:

y=x2=1

For x=9:

y=x2=7

The solutions are x=1 together with y=1 , and x=9 with y=7. Using coordinate pairs, the solutions are (1;1) and (9;7).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;1) and (9;7). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;1) and (9;7).


Submit your answer as: and

ID is: 3302 Seed is: 670

Simultaneous equations with xy terms

Determine the solution of the following equations:

3y=3xxy3x=y1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y1 in place of x in the hyperbolic equation and then solve for y.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation3y=3xxy3linear equationx=y1

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of x from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 3y=3xxy3 and substitute its value into x=y1. We could still get the same answer, but the calculations might become too long.
3y=3(y1)(y1)y30=3(y1)(y1)y3+3y0=3(y1)y(y1)3+3y0=y2+3y+y+3y330=y2+4y+3y330=y2+7y330=y2+7y6make the first term positivedivide each term by -1 to 0=y27y+6

STEP: Solve the equation 0=y27y+6
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y27y+60=(y6)(y1)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 6 and 1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

By using the values of y, we will calculate the corresponding values of x. We will do so by substituting each y-value into the linear equation.

NOTE: We can still get the same values of x if we substitute the values y into the equation 3y=3xxy3. If we do that, the calculations might become too long.

For y=6:

x=y1=5

For y=1:

x=y1=0

The solutions are x=5 together with y=6 , and x=0 with y=1. Using coordinate pairs, the solutions are (5;6) and (0;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (5;6) and (0;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (5;6) and (0;1).


Submit your answer as: and

ID is: 3302 Seed is: 414

Simultaneous equations with xy terms

Determine the solution of the following equations:

2y=7xxy+2y=2x+7
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x+7 in place of y in the hyperbolic equation and then solve for x.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=7xxy+2linear equationy=2x+7

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 2y=7xxy+2 and substitute its value into y=2x+7. We could still get the same answer, but the calculations might become too long.
2(2x+7)=7xx(2x+7)+20=7xx(2x+7)+22(2x+7)0=7x(2x2+7x)+2(4x+14)0=2x214x+4x+2140=2x210x+2140=2x210x12

STEP: Solve the equation 0=2x210x12
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x210x120=(x6)(x+1)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 6 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

By using the values of x, we will calculate the corresponding values of y. We will do so by substituting each x-value into the linear equation.

NOTE: We can still get the same values of y if we substitute the values x into the equation 2y=7xxy+2. If we do that, the calculations might become too long.

For x=6:

y=2x+7=5

For x=1:

y=2x+7=9

The solutions are x=6 together with y=5 , and x=1 with y=9. Using coordinate pairs, the solutions are (6;5) and (1;9).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (6;5) and (1;9). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (6;5) and (1;9).


Submit your answer as: and

3. Practical applications


ID is: 3280 Seed is: 601

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 42.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 42.
The numbers are consecutive numbers.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis42n1×n2=42

The correct equation for the first fact is n1×n2=42.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the second option.

The smaller numberis1 less than the larger numbern1=n21

This makes sense: consecutive numbers are separated by 1. The equation shows that with the 1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 42. n1×n2=42
The numbers are consecutive numbers. n1=n21

Submit your answer as: and

ID is: 3280 Seed is: 8952

Setting up simultaneous equations

Here are facts about two numbers:

  • The sum of the numbers is 19.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The sum of the numbers is 19.
The numbers are consecutive numbers.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The sum of the numbersis19n1+n2=19

The correct equation for the first fact is n1+n2=19.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The sum of the numbers is 19. n1+n2=19
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

ID is: 3280 Seed is: 7188

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 42.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 42.
The numbers are consecutive numbers.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis42n1×n2=42

The correct equation for the first fact is n1×n2=42.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 42. n1×n2=42
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

ID is: 381 Seed is: 1398

Word problems: distance, speed and time

Two helicopters start out at two different airports. The helicopters are 4,136 km apart. The helicopters start flying towards each other. One helicopter is flying at 242 km/h and the other helicopter at 275 km/h.

If both helicopters started their journey at the same time, how long will they take to pass each other?

Answer: The two helicopters pass each other after hours.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two helicopters must be equal to the total distance when the helicopters meet:

d1+d2=dtotald1+d2=4,136 km

STEP: Write equations to describe the motion of the helicopters
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the helicopters to meet. Let the time taken be t. Then we can write an expression for the distance each of the helicopters travels:

For helicopter 1:

d1=s1thelicopter is 242 km/hThe speed of the firstd1=242t

For helicopter 2:

d2=s2thelicopter is 275 km/hThe speed of the secondd2=275t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=4,136(242t)+(275t)=4,136517t=4,136t=4,136517t=8

The helicopters will meet after 8 hours.


Submit your answer as:

ID is: 381 Seed is: 7813

Word problems: distance, speed and time

Two ships start out at two different ports. The ships are 154 km apart. The ships start moving towards each other. One ship is moving at 88 km/h and the other ship at 66 km/h.

If both ships started their journey at the same time, how long will they take to pass each other?

Answer: The two ships pass each other after hours.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two ships must be equal to the total distance when the ships meet:

d1+d2=dtotald1+d2=154 km

STEP: Write equations to describe the motion of the ships
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the ships to meet. Let the time taken be t. Then we can write an expression for the distance each of the ships travels:

For ship 1:

d1=s1tship is 88 km/hThe speed of the firstd1=88t

For ship 2:

d2=s2tship is 66 km/hThe speed of the secondd2=66t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=154(88t)+(66t)=154154t=154t=154154t=1

The ships will meet after 1 hours.


Submit your answer as:

ID is: 381 Seed is: 5701

Word problems: distance, speed and time

Two cars start out at two different tollgates. The cars are 1,312 km apart. The cars start driving towards each other. One car is driving at 83 km/h and the other car at 81 km/h.

If both cars started their journey at the same time, how long will they take to pass each other?

Answer: The two cars pass each other after hours.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two cars must be equal to the total distance when the cars meet:

d1+d2=dtotald1+d2=1,312 km

STEP: Write equations to describe the motion of the cars
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the cars to meet. Let the time taken be t. Then we can write an expression for the distance each of the cars travels:

For car 1:

d1=s1tcar is 83 km/hThe speed of the firstd1=83t

For car 2:

d2=s2tcar is 81 km/hThe speed of the secondd2=81t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=1,312(83t)+(81t)=1,312164t=1,312t=1,312164t=8

The cars will meet after 8 hours.


Submit your answer as:

ID is: 392 Seed is: 17

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 7 pizzas and 6 hamburgers. Here are some facts about their lunch:

  • the total cost for the 7 pizzas and 6 hamburgers is R405
  • a pizza costs N=4 more than a hamburger

What is the price for one pizza and the price for one hamburger?

Answer:

A pizza costs N= and a hamburger costs N= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a pizza and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

p=the price of a pizzah=the price of a hamburger

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 7 pizzas and 6 hamburgers is N=405." We can use the expression 7p to represent the price of the 7 pizzas. Similarly, the expression 6h represents the price of the 6 hamburgers. With these values we can write a full equation for the prices:

In words: the total cost for the 7 pizzas and 6 hamburgers is N=405In maths: 7p+6h=405

Now we can use the second point. It says, "a pizza costs N=4 more than a hamburger." We can write this as an equation like this:

In words: a pizza costs N=4 more than a hamburgerIn maths: p=h+4

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

7p+6h=405p=h+4

We could use elimination, but substitution is a better choice (because p is already the subject). Substitute the second equation into the first equation and solve!

7p+6h=4057(h+4)+6h=4057h+28+6h=40513h=40528h=37713=29

This means that the price of one hamburger is N=29.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for h to find the value of p (the price of a pizza). We can use either equation to do this, but the second one is easier to use (because p is already isolated).

p=h+4=29+4=33

The price for one pizza is N=33.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the pizza is N=33 while a hamburger costs N=29.


Submit your answer as: and

ID is: 392 Seed is: 9913

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 7 salads and 8 milkshakes. Here are some facts about their lunch:

  • the total cost for the 7 salads and 8 milkshakes is R417
  • a salad costs N=6 more than a milkshake

What is the price for one salad and the price for one milkshake?

Answer:

A salad costs N= and a milkshake costs N= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a salad and a milkshake). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

s=the price of a saladm=the price of a milkshake

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 7 salads and 8 milkshakes is N=417." We can use the expression 7s to represent the price of the 7 salads. Similarly, the expression 8m represents the price of the 8 milkshakes. With these values we can write a full equation for the prices:

In words: the total cost for the 7 salads and 8 milkshakes is N=417In maths: 7s+8m=417

Now we can use the second point. It says, "a salad costs N=6 more than a milkshake." We can write this as an equation like this:

In words: a salad costs N=6 more than a milkshakeIn maths: s=m+6

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

7s+8m=417s=m+6

We could use elimination, but substitution is a better choice (because s is already the subject). Substitute the second equation into the first equation and solve!

7s+8m=4177(m+6)+8m=4177m+42+8m=41715m=41742m=37515=25

This means that the price of one milkshake is N=25.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for m to find the value of s (the price of a salad). We can use either equation to do this, but the second one is easier to use (because s is already isolated).

s=m+6=25+6=31

The price for one salad is N=31.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the salad is N=31 while a milkshake costs N=25.


Submit your answer as: and

ID is: 392 Seed is: 8556

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 4 sandwiches and 3 hamburgers. Here are some facts about their lunch:

  • the total cost for the 4 sandwiches and 3 hamburgers is R221
  • a sandwich costs N=8 more than a hamburger

What is the price for one sandwich and the price for one hamburger?

Answer:

A sandwich costs N= and a hamburger costs N= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a sandwich and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

s=the price of a sandwichh=the price of a hamburger

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 4 sandwiches and 3 hamburgers is N=221." We can use the expression 4s to represent the price of the 4 sandwiches. Similarly, the expression 3h represents the price of the 3 hamburgers. With these values we can write a full equation for the prices:

In words: the total cost for the 4 sandwiches and 3 hamburgers is N=221In maths: 4s+3h=221

Now we can use the second point. It says, "a sandwich costs N=8 more than a hamburger." We can write this as an equation like this:

In words: a sandwich costs N=8 more than a hamburgerIn maths: s=h+8

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

4s+3h=221s=h+8

We could use elimination, but substitution is a better choice (because s is already the subject). Substitute the second equation into the first equation and solve!

4s+3h=2214(h+8)+3h=2214h+32+3h=2217h=22132h=1897=27

This means that the price of one hamburger is N=27.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for h to find the value of s (the price of a sandwich). We can use either equation to do this, but the second one is easier to use (because s is already isolated).

s=h+8=27+8=35

The price for one sandwich is N=35.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the sandwich is N=35 while a hamburger costs N=27.


Submit your answer as: and

ID is: 3284 Seed is: 1357

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Gift buys some mangoes and some sweets. The total cost for 5 mangoes and 6 sweets is N=42.90. And each sweet costs N=1.10 more than each mango. What is the price for one mango?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be N=6.75. Or N=5. Or N=19.87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like N=4.30, or must a price always be a whole number like N=4.00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -N=5.50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3284 Seed is: 4898

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Riaan buys some mangoes and some packs of gum. The total cost for 5 mangoes and 2 packs of gum is N=33.90. And each pack of gum costs N=0.90 less than each mango. What is the price for one mango?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be N=6.75. Or N=5. Or N=19.87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like N=4.30, or must a price always be a whole number like N=4.00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -N=5.50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3284 Seed is: 528

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are fruits and sweets for sale. Ben buys some bananas and some chocolates. The total number of items he buys is 9. Each banana costs N=4.80 and each chocolate costs N=5.10. The total cost is N=44.70. How many bananas did Ben buy?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of items be a decimal number (or must it be an integer)?
  2. Can the number of items be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of bananas Ben buys. Can the number of items be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the number of items can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be N=6.75. Or N=5. Or N=19.87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the number of items can be a decimal number or not. In other words, can the number of bananas be a decimal number like 3.7 bananas? Or must it be a whole number like 4 bananas? This is different from the airtime example above: airtime can be a decimal number, but it is not possible to buy 3.5 bananas! (You might think, "Wait! I can buy 3.5 bananas, just cut one of them in half." But no one should expect to do this while shopping.)

The answer for the first question is: No, the number of items cannot be a decimal number.


STEP: Decide if the number of items can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the number of items can be negative, or must be postive. The number of bananas cannot be negative. If Ben did not buy any bananas at all, then the number of bananas is zero. But is it impossible for him to buy 3 bananas.

The answer to the second question is: No, the number of items cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3282 Seed is: 3478

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Atinuke is drawing a rectangle in her notebook. The length of the rectangle is 5 cm more than its width. The area of the rectangle is 6 cm2. What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the width of the rectangle be a decimal number (or must it be an integer)?
  2. Can the width of the rectangle be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176.3562 days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like 3.5 cm, or must a distance be a whole number like 4 cm? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have 3.5 days of school, but a rectangle certainly can have a width of 3.5 cm. Or 5.927 cm. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.


STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room 5 metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3282 Seed is: 8975

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Zainab is drawing a rectangle in her notebook. The length of the rectangle is 5 cm more than its width. The area of the rectangle is 6 cm2. What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the width of the rectangle be a decimal number (or must it be an integer)?
  2. Can the width of the rectangle be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176.3562 days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like 3.5 cm, or must a distance be a whole number like 4 cm? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have 3.5 days of school, but a rectangle certainly can have a width of 3.5 cm. Or 5.927 cm. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.


STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room 5 metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3282 Seed is: 8539

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Danjuma is drawing a pattern of rectangles in his notebook. The number of rectangles in each figure is 2 more than the number of rectangles in the previous figure. If there are 12 rectangles in the fourth figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of rectangles be a decimal number (or must it be an integer)?
  2. Can the number of rectangles be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Danjuma's diagram. Can the number of rectangles be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176.3562 days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like 3.7, or must the number be a whole number like 4? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. 3.7 rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get 3.7 rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.


STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Danjuma drew any rectangles at all, then the answer to this question must be positive. There is no such thing as 4 rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

ID is: 382 Seed is: 6307

Word problems: test scores and simultaneous equations

Fezekile and Atinuke are friends. Fezekile takes Atinuke's chemistry test paper and says: “I have 12 marks more than you do and the sum of both our marks is equal to 162. What are our marks?”

Answer:

Fezekile got marks and Atinuke got marks for the chemistry test paper.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

f= Fezekile's marka= Atinuke's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than AtinukeFezekile has 12 marksf=a+12(the total) is 162The sum of the marksf+a=162

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

f+a=162(a+12)+a=1622a=16212a=1502=75

This means that Atinuke's mark is 75.


STEP: Find Fezekile's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Atinuke's mark and one of the equations we have to find Fezekile's mark. The easiest way to do that is to substitute Atinuke's mark back into the first equation:

f=a+12=(75)+12=87

The students achieved these marks: Fezekile earned 87 and Atinuke earned 75.


Submit your answer as: and

ID is: 382 Seed is: 6703

Word problems: test scores and simultaneous equations

Kamogelo and Danjuma are friends. Kamogelo takes Danjuma's electrical technology test paper and says: “I have 7 marks more than you do and the sum of both our marks is equal to 161. What are our marks?”

Answer:

Kamogelo got marks and Danjuma got marks for the electrical technology test paper.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

k= Kamogelo's markd= Danjuma's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than DanjumaKamogelo has 7 marksk=d+7(the total) is 161The sum of the marksk+d=161

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

k+d=161(d+7)+d=1612d=1617d=1542=77

This means that Danjuma's mark is 77.


STEP: Find Kamogelo's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Danjuma's mark and one of the equations we have to find Kamogelo's mark. The easiest way to do that is to substitute Danjuma's mark back into the first equation:

k=d+7=(77)+7=84

The students achieved these marks: Kamogelo earned 84 and Danjuma earned 77.


Submit your answer as: and

ID is: 382 Seed is: 9159

Word problems: test scores and simultaneous equations

Kungawo and Amarachi are friends. Kungawo takes Amarachi's chemistry test paper and says: “I have 8 marks more than you do and the sum of both our marks is equal to 132. What are our marks?”

Answer:

Kungawo got marks and Amarachi got marks for the chemistry test paper.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

k= Kungawo's marka= Amarachi's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than AmarachiKungawo has 8 marksk=a+8(the total) is 132The sum of the marksk+a=132

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

k+a=132(a+8)+a=1322a=1328a=1242=62

This means that Amarachi's mark is 62.


STEP: Find Kungawo's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Amarachi's mark and one of the equations we have to find Kungawo's mark. The easiest way to do that is to substitute Amarachi's mark back into the first equation:

k=a+8=(62)+8=70

The students achieved these marks: Kungawo earned 70 and Amarachi earned 62.


Submit your answer as: and

ID is: 3278 Seed is: 9271

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The sum of the numbers is 14.

What is the value of the larger number?

Answer: The larger number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of 14. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 14". Remember that sum means addition. So:

n1+n2=14

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

We can solve this using substitution. However, remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)+n2=142n22=142n2=16n2=8

The result is n2=8. Notice that this means that the other number, n1, must be 6, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 14, and 8+6=14.

The larger number is 8.


Submit your answer as:

ID is: 3278 Seed is: 4244

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The sum of the numbers is 18.

What is the value of the larger number?

Answer: The larger number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of 18. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 18". Remember that sum means addition. So:

n1+n2=18

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

We can solve this using substitution. However, remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)+n2=182n22=182n2=20n2=10

The result is n2=10. Notice that this means that the other number, n1, must be 8, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 18, and 10+8=18.

The larger number is 10.


Submit your answer as:

ID is: 3278 Seed is: 5643

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The sum of the numbers is 18.

Find the value of the smaller number.

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of 18. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 18". Remember that sum means addition. So:

n1+n2=18

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=18, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=182n1+2=182n1=16n1=8

The result is n1=8. Notice that this means that the other number, n2, must be 10, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 18, and 8+10=18.

The smaller number is 8.


Submit your answer as:

ID is: 3286 Seed is: 136

Word problems: checking answers

Zainab is 11 years older than her sister, Nthabiseng. In 5 years, Zainab will be 2 times as old as Nthabiseng. How old is Nthabiseng now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Nthabiseng is years old.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Nthabiseng's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Nthabiseng's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Zainab and Nthabiseng. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aZ=Zainab's ageaN=Nthabiseng's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Zainab is 11 years older than her sister, Nthabiseng." We need to translate that into mathematics. The key word is older, which tells us to use addition to relate the ages.

aZ=aN+11

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 5 years, Zainab will be 2 times as old as Nthabiseng." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aZ+5=Zainab's age in 5 yearsaN+5=Nthabiseng's age in 5 years

These are the ages at which Zainab will be 2 times as old as Nthabiseng. We can put all this together as follows:

in 5 yearsZainab's age=2×in 5 yearsNthabiseng's ageaZ+5=2(aN+5)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aZ and aN. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aZ=aN+11. We can substitute this into the second equation and solve for aN.

aZ+5=2(aN+5)(aN+11)+5=2(aN+5)aN+16=2aN+106=aN

Terrific: we have the answer.

Nthabiseng is 6 years old.

The correct answer is: 6.


Submit your answer as:

ID is: 3286 Seed is: 1416

Word problems: checking answers

Lisa is 16 years younger than her brother, Babatunde. In 5 years, Babatunde will be 5 times as old as Lisa. How old is Lisa now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Lisa is years old.

one-of
type(string.nocase)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Lisa's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Lisa's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Babatunde and Lisa. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aB=Babatunde's ageaL=Lisa's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Lisa is 16 years younger than her brother, Babatunde." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aL=aB16

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 5 years, Babatunde will be 5 times as old as Lisa." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aB+5=Babatunde's age in 5 yearsaL+5=Lisa's age in 5 years

These are the ages at which Babatunde will be 5 times as old as Lisa. We can put all this together as follows:

in 5 yearsBabatunde's age=5×in 5 yearsLisa's ageaB+5=5(aL+5)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aB and aL. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aL=aB16. We can substitute this into the second equation and solve for aB.

aB+5=5(aL+5)aB+5=5((aB16)+5)aB+5=5aB5560=4aB15=aB

Terrific: this means that Babatunde is 15 years old. But the question asked for us to find Lisa's age. We can find it using the first equation, which relates the two ages:

aL=aB16=(15)16=1

So we finally got the answer to the question. But wait a minute: aL represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.


Submit your answer as:

ID is: 3286 Seed is: 8752

Word problems: checking answers

Kelly is 10 years younger than her sister, Chinyelu. In 6 years, Chinyelu will be 2 times as old as Kelly. How old is Kelly now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Kelly is years old.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Kelly's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Kelly's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Chinyelu and Kelly. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aC=Chinyelu's ageaK=Kelly's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Kelly is 10 years younger than her sister, Chinyelu." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aK=aC10

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 6 years, Chinyelu will be 2 times as old as Kelly." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aC+6=Chinyelu's age in 6 yearsaK+6=Kelly's age in 6 years

These are the ages at which Chinyelu will be 2 times as old as Kelly. We can put all this together as follows:

in 6 yearsChinyelu's age=2×in 6 yearsKelly's ageaC+6=2(aK+6)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aC and aK. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aK=aC10. We can substitute this into the second equation and solve for aC.

aC+6=2(aK+6)aC+6=2((aC10)+6)aC+6=2aC814=aC

Terrific: this means that Chinyelu is 14 years old. But the question asked for us to find Kelly's age. We can find it using the first equation, which relates the two ages:

aK=aC10=(14)10=4

So we finally got the answer to the question. Kelly is 4 years old.

The correct answer is: 4.


Submit your answer as:

ID is: 3279 Seed is: 8309

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The product of the numbers is 24.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a product of 24. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 24". Remember that product means multiplication. So:

n1n2=24

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=24, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=24n12+2n1=24n12+2n124=0(n1+6)(n14)=0
n1=6andn1=4

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=4.

This means the other number, n2, must be 6, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 24, and 46=24.

The smaller number is 4.


Submit your answer as:

ID is: 3279 Seed is: 5324

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 35.

Determine the value of the smaller number.

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 35. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 35". Remember that product means multiplication. So:

n1n2=35

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=35, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=35n12+2n1=35n12+2n135=0(n1+7)(n15)=0
n1=7andn1=5

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=5.

This means the other number, n2, must be 7, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 35, and 57=35.

The smaller number is 5.


Submit your answer as:

ID is: 3279 Seed is: 4438

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive even integers.
  • The product of the numbers is 120.

Determine the value of the smaller number.

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a product of 120. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, 10 and 12 are consecutive even numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the odd integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 120". Remember that product means multiplication. So:

n1n2=120

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=120, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=120n12+2n1=120n12+2n1120=0(n1+12)(n110)=0
n1=12andn1=10

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=10.

This means the other number, n2, must be 12, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 120, and 1012=120.

The smaller number is 10.


Submit your answer as:

ID is: 380 Seed is: 6626

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are positive
  • the product of the numbers is 342

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 342.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 342. That means if we multiply the numbers we get 342.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 342. In other words, if we multiply the numbers, we get 342. As an equation this is:

n1n2=342

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=342n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=342(n1)2+n1342=0(n1+19)(n118)=0
n1=19andn1=18

The solutions to the equation are n1=19 or n1=18. This means the first number is either −19 or 18.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −19 and 18 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the n1=19, so the first number is n1=18.

Based on that we can find the second number. The second number is n1+1, which is equal to (18)+1=19.

The two consecutive integers are 18 and 19.


Submit your answer as: and

ID is: 380 Seed is: 9370

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 20

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 20.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 20. That means if we multiply the numbers we get 20.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 20. In other words, if we multiply the numbers, we get 20. As an equation this is:

n1n2=20

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=20n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=20(n1)2+n120=0(n1+5)(n14)=0
n1=5andn1=4

The solutions to the equation are n1=5 or n1=4. This means the first number is either −5 or 4.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −5 and 4 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=4, so the first number is n1=5.

Based on that we can find the second number. The second number is n1+1, which is equal to (5)+1=4.

The two consecutive integers are −5 and −4.


Submit your answer as: and

ID is: 380 Seed is: 9700

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 30

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 30.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 30. That means if we multiply the numbers we get 30.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 30. In other words, if we multiply the numbers, we get 30. As an equation this is:

n1n2=30

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=30n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=30(n1)2+n130=0(n1+6)(n15)=0
n1=6andn1=5

The solutions to the equation are n1=6 or n1=5. This means the first number is either −6 or 5.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −6 and 5 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=5, so the first number is n1=6.

Based on that we can find the second number. The second number is n1+1, which is equal to (6)+1=5.

The two consecutive integers are −6 and −5.


Submit your answer as: and

ID is: 385 Seed is: 7416

Word problems: an age-old question

Ekene has a son, Nkosingiphile. Here are some facts about how old Ekene and Nkosingiphile are:

  • Ekene is 4 times as old as Nkosingiphile right now.
  • 8 years from now, Ekene will be 3 times as old as Nkosingiphile.

How old are Ekene and Nkosingiphile now?

Answer:

Ekene is years old and Nkosingiphile is years old.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Ekene and his son, Nkosingiphile. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let e=Ekene's ageLet n=Nkosingiphile's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Ekene is now 4 times as old as Nkosingiphile." As an equation, this is:

Ages now: e=4n

The second piece of information says that in "8 years... Ekene will be 3 times as old as his son." In 8 years Ekene will be e+8 years old, and similarly Nkosingiphile will be n+8 years old. Then:

Ages in 8 years: e+8=3(n+8)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel e.)

e+8=3(n+8)(4n)+8=3n+24n=16

Great! Now we know that Nkosingiphile is 16 years old.


STEP: Use Nkosingiphile's age to find Ekene's age
[−1 point ⇒ 0 / 6 points left]

We can now find Ekene's age. Substitute Nkosingiphile's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

e=4n=4(16)=64

Write your final answer: Ekene is 64 years old and Nkosingiphile is 16 years old.


Submit your answer as: and

ID is: 385 Seed is: 7439

Word problems: an age-old question

Chukwuma has a son, Nqobani. Here are some facts about how old Chukwuma and Nqobani are:

  • Chukwuma is 4 times as old as Nqobani right now.
  • 7 years from now, Chukwuma will be 3 times as old as Nqobani.

How old are Chukwuma and Nqobani now?

Answer:

Chukwuma is years old and Nqobani is years old.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Chukwuma and his son, Nqobani. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let c=Chukwuma's ageLet n=Nqobani's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Chukwuma is now 4 times as old as Nqobani." As an equation, this is:

Ages now: c=4n

The second piece of information says that in "7 years... Chukwuma will be 3 times as old as his son." In 7 years Chukwuma will be c+7 years old, and similarly Nqobani will be n+7 years old. Then:

Ages in 7 years: c+7=3(n+7)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel c.)

c+7=3(n+7)(4n)+7=3n+21n=14

Great! Now we know that Nqobani is 14 years old.


STEP: Use Nqobani's age to find Chukwuma's age
[−1 point ⇒ 0 / 6 points left]

We can now find Chukwuma's age. Substitute Nqobani's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

c=4n=4(14)=56

Write your final answer: Chukwuma is 56 years old and Nqobani is 14 years old.


Submit your answer as: and

ID is: 385 Seed is: 4626

Word problems: an age-old question

Talwar has a son, Delphino. Here are some facts about how old Talwar and Delphino are:

  • Talwar is 10 times as old as Delphino right now.
  • 10 years from now, Talwar will be 4 times as old as Delphino.

How old are Talwar and Delphino now?

Answer:

Talwar is years old and Delphino is years old.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Talwar and his son, Delphino. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let t=Talwar's ageLet d=Delphino's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Talwar is now 10 times as old as Delphino." As an equation, this is:

Ages now: t=10d

The second piece of information says that in "10 years... Talwar will be 4 times as old as his son." In 10 years Talwar will be t+10 years old, and similarly Delphino will be d+10 years old. Then:

Ages in 10 years: t+10=4(d+10)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel t.)

t+10=4(d+10)(10d)+10=4d+406d=30d=5

Great! Now we know that Delphino is 5 years old.


STEP: Use Delphino's age to find Talwar's age
[−1 point ⇒ 0 / 6 points left]

We can now find Talwar's age. Substitute Delphino's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

t=10d=10(5)=50

Write your final answer: Talwar is 50 years old and Delphino is 5 years old.


Submit your answer as: and

ID is: 3285 Seed is: 5149

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42.195 kilometres. 1,000 people ran the "Sunshine and Roses are for Sissies" marathon in Egoli. The second-place runner finished 15.51 seconds after the winner. The third runner was 13.44 seconds behind the second runner, and the fourth runner was 28.31 seconds behind the third runner. If the second-place runner had an average speed of 18.57 km/h, what was the winning time for the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. Negative time is not possible: it must be a positive number.

The correct answer is: No, it cannot be negative.


Submit your answer as:

ID is: 3285 Seed is: 9468

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42.195 kilometres. 1,000 people ran the "One Metre at a Time" marathon in Cape Town. The second-place runner finished 13.45 seconds after the winner. The third runner was 16.97 seconds behind the second runner, and the fourth runner was 20.67 seconds behind the third runner. If the second-place runner had an average speed of 19.39 km/h, how many of the runners took more than 2.1 hours to finish the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: how many of the runners took more than 2.1 hours to finish the marathon. So the question is about the number of people who finished the race after 2.1 hours had passed. The answer cannot be negative. The number of people to finish after 2.1 hours might be 3 or 7 people, but it cannot be 5 people. In other words, the number of people must be a positive number or zero, but it cannot be negative.

The correct answer is: No, it cannot be negative.


Submit your answer as:

ID is: 3285 Seed is: 6771

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42.195 kilometres. 1,000 people ran the "It Ain't Over Till It's Over" marathon in Polokwane. The second-place runner finished 11.65 seconds after the winner. The third runner was 14.36 seconds behind the second runner, and the fourth runner was 25.81 seconds behind the third runner. If the second-place runner had an average speed of 19.18 km/h, what was the winning time for the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. Negative time is not possible: it must be a positive number.

The correct answer is: No, it cannot be negative.


Submit your answer as:

ID is: 377 Seed is: 9759

Word problems: rectangle facts

The diagonal of a rectangle is 34 cm more than its width. The length of the same rectangle is 17 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is 34 cm more than its width:

d=w+34

The length of the rectangle is 17 cm more than its width:

l=w+17

STEP: Use the theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+34 in for d and w+17 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+34)2=w2+(w+17)2w2+68w+1,156=w2+(w2+34w+289)0=w234w867

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w234w8670=(w51)(w+17)w=51or w=17

This means that the width of the rectangle is either 51 cm or 17 cm. But the dimensions of a rectangle cannot be negative, so w=51 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+17=(51)+17=68 cm

The width of the rectangle is 51 cm and the length is 68 cm.


Submit your answer as: and

ID is: 377 Seed is: 2823

Word problems: rectangle facts

The diagonal of a rectangle is 26 cm more than its width. The length of the same rectangle is 13 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is 26 cm more than its width:

d=w+26

The length of the rectangle is 13 cm more than its width:

l=w+13

STEP: Use the theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+26 in for d and w+13 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+26)2=w2+(w+13)2w2+52w+676=w2+(w2+26w+169)0=w226w507

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w226w5070=(w39)(w+13)w=39or w=13

This means that the width of the rectangle is either 39 cm or 13 cm. But the dimensions of a rectangle cannot be negative, so w=39 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+13=(39)+13=52 cm

The width of the rectangle is 39 cm and the length is 52 cm.


Submit your answer as: and

ID is: 377 Seed is: 8983

Word problems: rectangle facts

The diagonal of a rectangle is 30 cm more than its width. The length of the same rectangle is 15 cm more than its width.

What are the width and the length of the rectangle?

Answer: The width is cm and length is cm.
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is 30 cm more than its width:

d=w+30

The length of the rectangle is 15 cm more than its width:

l=w+15

STEP: Use the theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+30 in for d and w+15 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+30)2=w2+(w+15)2w2+60w+900=w2+(w2+30w+225)0=w230w675

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w230w6750=(w45)(w+15)w=45or w=15

This means that the width of the rectangle is either 45 cm or 15 cm. But the dimensions of a rectangle cannot be negative, so w=45 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+15=(45)+15=60 cm

The width of the rectangle is 45 cm and the length is 60 cm.


Submit your answer as: and

ID is: 3281 Seed is: 3660

Setting up simultaneous equations

Last week, Amaka and Bukelwa had a chemistry test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 148.
  • Amaka's mark is 18 less than Bukelwa's mark.

Let a represent Amaka's mark and b represent Bukelwa's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 148.
Amaka's mark is 18 less than Bukelwa's mark.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Amaka's mark and b for Bukelwa's mark. So we can break up the first fact like this:

The sum of the marksis148a+b=148

The first fact is equivalent to this equation: a+b=148.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Amaka's markis18 less than Bukelwa's marka=b18

This equation, a=b18, means that Amaka's mark is less that Bukelwa's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 148. a+b=148
Amaka's mark is 18 less than Bukelwa's mark. a=b18

Submit your answer as: and

ID is: 3281 Seed is: 5232

Setting up simultaneous equations

Last week, Anthea and Balarba had a chemistry test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 137.
  • Anthea's mark is 5 more than Balarba's mark.

Let a represent Anthea's mark and b represent Balarba's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 137.
Anthea's mark is 5 more than Balarba's mark.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Anthea's mark and b for Balarba's mark. So we can break up the first fact like this:

The sum of the marksis137a+b=137

The first fact is equivalent to this equation: a+b=137.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Anthea's markis5 more than Balarba's marka=b+5

This equation, a=b+5, means that Anthea's mark is more that Balarba's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 137. a+b=137
Anthea's mark is 5 more than Balarba's mark. a=b+5

Submit your answer as: and

ID is: 3281 Seed is: 5750

Setting up simultaneous equations

Last week, Adaeze and Balarabe had a chemistry test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 115.
  • Adaeze's mark is 5 less than Balarabe's mark.

Let a represent Adaeze's mark and b represent Balarabe's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 115.
Adaeze's mark is 5 less than Balarabe's mark.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Adaeze's mark and b for Balarabe's mark. So we can break up the first fact like this:

The sum of the marksis115a+b=115

The first fact is equivalent to this equation: a+b=115.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Adaeze's markis5 less than Balarabe's marka=b5

This equation, a=b5, means that Adaeze's mark is less that Balarabe's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 115. a+b=115
Adaeze's mark is 5 less than Balarabe's mark. a=b5

Submit your answer as: and